variance is given as 100.
Hence, standard deviation is 10. Mean = 270.
For the women to have a long pregenancy more than 300 days or less than 240 days.Here depending upon the defintion of long and short pregenany there answer may vary. As we know mean is 270 and standard deviation is 10. this range specified $\mu + 3\sigma and \mu -3\sigma$v i.e that the suspect must have been out of country in that range. So answer should 0.996. It must be true that he should've been out of country at that range. May be if they had asked negation of that question then it will be 0.004. I don't know how its 0.241.
For her to have a long pregenancy the suspect must've been within the country300 days before pregenancy. And for her to have a short pregnancy he must've been within the country 240 days before pregnancy.